I apologize if this isn't the appropriate place to ask this. Anyone know what's going on with Project Euler? I try to access their forum and get a certificate error telling me I should close the site immediately. The only contact info they give is through the forum. They have an email for new problems, but they've stopped posting problems until the fall.
I'm half-considering going to a problem I've completed, making a post in the solution thread about what's going on and reporting it if this doesn't clear up.
Py = log(y)
Px = log(x)
y = f(x)
x = e^Px
Py = log(f(x)) = log(f(e^Px))
if f(x) = x
Py = log(e^Px) = Px
so f(x) = x would give a straight line
yet I was reading this book that described a power-law distribution as something that would be plotted as a straight line in this kind of graph
but f(x) = x doesn't seem to have anything to do with powers
oh but way, maybe as you say, x² will also give a straight line
oh right, x² would give 2 Px
I see, thanks :)
another bit of confusion is that it sounds like when I plot log(e^log(x)) on a same-interval scale, I should still get a straight line, as if mimicking the logarithmic version of y=x
it depends on your perspective, but i like topological spaces where the topology has something to do with the setting (beyond being just compatible with the maps you need or some algebraic operations)
for example the fibres of the etale space are all discrete
it follows from $K(H)$ having no self-adjoint two-sided ideals, as the kernel of a character must be such an ideal
if you suppose you have such an ideal and take a self-adjoint element and conjugate by the appropriate projection you get a rank one map, the associated projection must lie in the ideal, by conjugating with appropriate rank 2 maps you can assume any rank 1 projection is in the ideal
the span of the rank 1 projections is dense in $K(H)$ however
(and two sided ideals are closed in $C^*$ algebras, although the kernel of a character is also automatically closed)
How much difference is there between those sets of operators when H is finite-dimensional? My knee-jerk reaction would be that there can’t be much difference
Any locally constant $G$-valued sheaf on $X$ is constant iff $X$ is simply connected. Because the hypothesis is equivalent to demanding any $G$-representation of $\pi_1(X)$ is trivial.
I mean it's not that cool, really, if you think about it. Any space $X$ has a locally constant $\pi_1(X)$-valued (set-valued) sheaf on it, given by the left-regular representation of $\pi_1(X)$. The espace etale is the universal cover if $X$ was a sufficiently nice space.
@Balarka apparently, there are some countable groups with no finite-dimensional $\Bbb R$-linear representation. This means that bundles with flat connection on a smooth manifold can't detect if the manifold is simply connected
there are even finitely presented examples, so even assuming compactness doesn't help
but if we assume a compact 3-manifold, then flat connections do detect simply-connectedness: any compact 3-manifold group is residually finite, so it admits a non-trivial homomorphism to some finite group, we can compose that with a faithful representation of that finite group
Choose a presentation of $G$, and take a wedge of $n$-spheres indexed by the generators of $G$, and then attach $n+1$-cells indexed by the relators. Now glue higher cells to kill higher homotopy groups.
If $X$ is a CW complex, with a chosen class $\alpha \in \pi_n X$, then consider a representative $S^n \to X$ of this class. By cellular approximation theorem make it a cellular map. Then glue a $D^{n+1}$ by considering this to be your attaching map.
You say in the resulting space $X'$ you have "killed $\alpha$"
This partially explains your problem as well. Consider instead $S^2 \vee S^1$. It's universal cover is homotopy equivalent to an infinite-wedge of $S^2$'s, so their higher homotopy groups are really really unpleasant compared to that of $S^2$ or $S^1$.
@LeakyNun It appears the problem has been corrected. The site itself was fine. The problem was with the forums. Both old bookmarks and the link to projecteuler.chat from the About page on projecteuler.net had the issue. It now loads fine with no issues.
@MatheinBoulomenos It's the algebraic proof of Hopf degree theorem, really. If you have two maps $f, g : M \to S^n$ with same degree, then they give rise to equal classes $H^n(f) = H^n(g)$ in $H^n(M; \Bbb Z) = [M, K(\Bbb Z, n)]$. By a Yoneda-type argument you can argue that everything is natural and the corresponding maps $f, g : M \to S^n \subset K(\Bbb Z, n)$ are homotopic.
@BalarkaSen At least you get Atiyah-Hirzebruch $H_*(X;\pi_*^s) \implies \pi_*^s(X)$ coming from the cellular filtration but obviously it's very complicated to pull out the differentials, much less solve the filtration problems in the end
